Question

Is it true that if f:R→S is a ring homomorphism between two rings with unity R and S then f+1; defined by (f+1) (r) =f(r) +1 is also a ring homeomorphism from R to S?

Answer 1

Let $f:R→S$ is a ring homomorphism between two rings with unity $R$ and $S$ . Consider the map$f+1: R\to S, (f+1) (r) =f(r) +1.$

Since $f$ is a homomorphism, $(f+1)(r+s)=f(r+s)+1=f(r)+f(s)+1$. Since $S$ is Abelian group according to addition operation, we conclude that

$(f+1)(r)+(f+1)(s)=f(r)+1+f(s)+1=f(r)+f(s)+1+1$. If $(f+1)(r+s)=(f+1)(r)+(f+1)(s)$, then $f(r)+f(s)+1=f(r)+f(s)+1+1$. Using left cancelation law in Abelian group of a ring, we obtain that $1=0$, and then $s=s\cdot 1=s\cdot 0=0$ for any $s\in S$. Therefore, if $f+1: R\to S$ is a ring homomorphism, then $S=\{0\}$ is a trivial ring. If $S$ is a non-trivial ring, then $f+1: R\to S$ is not a homomorphism.

Answer: false